{"id":6901,"date":"2011-06-07T00:11:50","date_gmt":"2011-06-06T23:11:50","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=6901"},"modified":"2022-01-10T22:29:41","modified_gmt":"2022-01-10T22:29:41","slug":"resolve-as-equacoes-utilizando-a-lei-do-anulamento-do-produto","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=6901","title":{"rendered":"Resolve as equa\u00e7\u00f5es, utilizando a lei do anulamento do produto"},"content":{"rendered":"<p><ul id='GTTabs_ul_6901' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_6901' class='GTTabs_curr'><a  id=\"6901_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_6901' ><a  id=\"6901_1\" onMouseOver=\"GTTabsShowLinks('Resolu\u00e7\u00e3o'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Resolu\u00e7\u00e3o<\/a><\/li>\n<\/ul>\n\n<div class='GTTabs_divs GTTabs_curr_div' id='GTTabs_0_6901'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Resolve as equa\u00e7\u00f5es, utilizando a lei do anulamento do produto:<\/p>\n<ol>\n<li>$x(x+2)=0$<\/li>\n<li>$(2x+1)(x-\\frac{1}{3})=0$<\/li>\n<li>${{x}^{2}}+3x=0$<\/li>\n<li>$3{{z}^{2}}-12z=0$<\/li>\n<li>$(x-3)(2+7x)=0$<\/li>\n<li>$x(x+1)+2(x+1)=0$<\/li>\n<li>$-x(x+4)=0$<\/li>\n<li>$(x+4)x-3(x+4)=0$<\/li>\n<li>$3(x-2)(x+2)=0$<\/li>\n<li>$16x+2{{x}^{2}}=0$<\/li>\n<li>$2{{m}^{2}}+5m=0$<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_6901' onClick='GTTabs_show(1,6901)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_6901'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<blockquote>\n<p style=\"text-align: center;\"><strong>Lei do anulamento do produto<\/strong><\/p>\n<p style=\"text-align: center;\">Um produto \u00e9 nulo se e s\u00f3 se pelo menos um dos fatores for nulo.<\/p>\n<h3 style=\"text-align: center;\">$\\begin{matrix}<br \/>\nA\\times B=0 &amp; \\Leftrightarrow &amp; A=0\\vee B=0 \\\\<br \/>\n\\end{matrix}$<\/h3>\n<\/blockquote>\n<ol>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\nx(x+2)=0 &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nx=0 &amp; \\vee\u00a0 &amp; x+2=0\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nx=0 &amp; \\vee\u00a0 &amp; x=-2\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n(2x+1)(x-\\frac{1}{3})=0 &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\n2x+1=0 &amp; \\vee\u00a0 &amp; x-\\frac{1}{3}=0\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nx=-\\frac{1}{2} &amp; \\vee\u00a0 &amp; x=\\frac{1}{3}\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n{{x}^{2}}+3x=0 &amp; \\Leftrightarrow\u00a0 &amp; x(x+3)=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nx=0 &amp; \\vee\u00a0 &amp; x+3=0\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nx=0 &amp; \\vee\u00a0 &amp; x=-3\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n3{{z}^{2}}-12z=0 &amp; \\Leftrightarrow\u00a0 &amp; 3z(z-4)=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\n3z=0 &amp; \\vee\u00a0 &amp; z-4=0\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nz=0 &amp; \\vee\u00a0 &amp; z=4\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n(x-3)(2+7x)=0 &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nx-3=0 &amp; \\vee\u00a0 &amp; 2+7x=0\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nx=3 &amp; \\vee\u00a0 &amp; x=-\\frac{2}{7}\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\nx(x+1)+2(x+1)=0 &amp; \\Leftrightarrow\u00a0 &amp; (x+1)(x+2)=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nx+1=0 &amp; \\vee\u00a0 &amp; x+2=0\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nx=-1 &amp; \\vee\u00a0 &amp; x=-2\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n-x(x+4)=0 &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\n-x=0 &amp; \\vee\u00a0 &amp; x+4=0\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nx=0 &amp; \\vee\u00a0 &amp; x=-4\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n(x+4)x-3(x+4)=0 &amp; \\Leftrightarrow\u00a0 &amp; (x+4)(x-3)=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nx+4=0 &amp; \\vee\u00a0 &amp; x-3=0\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nx=-4 &amp; \\vee\u00a0 &amp; x=3\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n3(x-2)(x+2)=0 &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nx-2=0 &amp; \\vee\u00a0 &amp; x+2=0\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nx=2 &amp; \\vee\u00a0 &amp; x=-2\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n16x+2{{x}^{2}}=0 &amp; \\Leftrightarrow\u00a0 &amp; 2x(8+x)=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\n2x=0 &amp; \\vee\u00a0 &amp; 8+x=0\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nx=0 &amp; \\vee\u00a0 &amp; x=-8\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n2{{m}^{2}}+5m=0 &amp; \\Leftrightarrow\u00a0 &amp; m(2m+5)=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nm=0 &amp; \\vee\u00a0 &amp; 2m+5=0\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; \\begin{array}{*{35}{l}}<br \/>\nm=0 &amp; \\vee\u00a0 &amp; m=-\\frac{5}{2}\u00a0 \\\\<br \/>\n\\end{array}\u00a0 \\\\<br \/>\n\\end{array}\\]<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_6901' onClick='GTTabs_show(0,6901)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Resolve as equa\u00e7\u00f5es, utilizando a lei do anulamento do produto: $x(x+2)=0$ $(2x+1)(x-\\frac{1}{3})=0$ ${{x}^{2}}+3x=0$ $3{{z}^{2}}-12z=0$ $(x-3)(2+7x)=0$ $x(x+1)+2(x+1)=0$ $-x(x+4)=0$ $(x+4)x-3(x+4)=0$ $3(x-2)(x+2)=0$ $16x+2{{x}^{2}}=0$ $2{{m}^{2}}+5m=0$ Resolu\u00e7\u00e3o &gt;&gt; Resolu\u00e7\u00e3o &lt;&lt; Enunciado<\/p>\n","protected":false},"author":1,"featured_media":14083,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[100,97,193],"tags":[198],"series":[],"class_list":["post-6901","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-8--ano","category-aplicando","category-equacoes-de-grau-superior-ao-1-","tag-lei-do-anulamento-do-produto"],"views":41291,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/03\/Mat28.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/6901","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=6901"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/6901\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/14083"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=6901"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=6901"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=6901"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=6901"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}